1. Copyright © 2008 by Hawkes Learning Systems/Quant Systems, Inc.
All rights reserved.
HAWKES LEARNING SYSTEMS
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Section 11.6
ANOVA

2. ANOVA – analysis of variance.
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Hypothesis Testing (Two or More Populations)
11.6 ANOVA
Definitions:
ANOVA – analysis of variance.
Grand Mean – the weighted mean of the sample means from each of the populations.
Total Variation – the sum of the variations contributed by each sample.
Sum of Squares among Treatments, SST – the variation resulting from the differences in the population means.
Sum of Squares of Error, SSE – the variation resulting from the variance within the populations.

3. The distributions of all of the populations are approximately normal.
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Hypothesis Testing (Two or More Populations)
11.6 ANOVA
ANOVA Test:
The distributions of all of the populations are approximately normal.
The variances of the populations are the same. (If the sample sizes are nearly equal, this assumption is not essential.)
Independent, simple random samples were taken from each population.

4. Ha: At least one mean differs from the others.
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Hypothesis Testing (Two or More Populations)
11.6 ANOVA
Null and Alternative Hypotheses:
H0: 1 = 2 = … = n.
Ha: At least one mean differs from the others.
An ANOVA test indicates if at least one of the population means is different, but it DOES NOT indicate which one differs, or by how much.

5. xij = the j th data value from population i
HAWKES LEARNING SYSTEMS
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Hypothesis Testing (Two or More Populations)
11.6 ANOVA
Grand Mean:
xij = the j th data value from population i
ni = the sample size from population i
∑ xij = the sum of the data values in the i th population
Example: Treatments A , B and C A= B= C=
Total Average = (Grand Mean)

6. SS DF MS F Treatments Error Total
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Hypothesis Testing (Two or More Populations)
11.6 ANOVA
ANOVA Table: SS DF MS F
Treatments
SST
k – 1
Error
SSE
n – k
Total
SST + SSE
DFT + DFE

7. k = the number of populations
HAWKES LEARNING SYSTEMS
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Hypothesis Testing (Two or More Populations)
11.6 ANOVA
Sum of Squares among Treatments: the variation resulting from the
differences in the population means.
k = the number of populations
ni = the sample size from the i th population
= the sample mean from the i th population
Example: Treatments A , B and C
A= Average = 5.89 n=9
B= Average = 5.67 n=9
C= Average = 5.11 n=9
Total Average = (Grand Mean)
SST= 9*(5.89 – 5.55)2 + 9*(5.67 – 5.55)2 + 9*(5.11 – 5.55)2
SST = =

8. HAWKES LEARNING SYSTEMS
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Hypothesis Testing (Two or More Populations)
11.6 ANOVA
Sum of Squares of Error: the variation resulting from the variance
within the populations.
Example: Treatments A , B and C
A= Average = 5.89 n=9
B= Average = 5.67 n=9
C= Average = 5.11 n=9
SSE(A) = (2-5.89)2 + (3-5.89)2 + …… + ( )2 =
SSE(B) = (3-5.67)2 + (5-5.67)2 + …… + (9-5.67)2 = 74
SSE(C) = (5-5.11)2 + (6-5.11)2 + …… + (8-5.11)2 =
SSE = SSE(A) + SSE(B) + SSE(C) =

9. Total Variation = SST + SSE Total Variation = 2.888889 + 167.7778
HAWKES LEARNING SYSTEMS
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Hypothesis Testing (Two or More Populations)
11.6 ANOVA
Total Variation: the sum of the variations contributed by each sample.
Total Variation = SST + SSE
Total Variation =
Total Variation =
Example: Treatments A , B and C
A= Average = 5.89 n=9
B= Average = 5.67 n=9
C= Average = 5.11 n=9
Total Average = (Grand Mean)
SSTotal = (2-5.55)2 + (3-5.55)2 + …… + (7-5.55)2 + (8-5.55)2 =

10. MST = 2.888889 ÷ 2 = 1.44444 HAWKES LEARNING SYSTEMS
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Hypothesis Testing (Two or More Populations)
11.6 ANOVA
Mean Square for Treatment: =
k = number of treatments = 3
Degrees of Freedom of Treatments (DFT) = k -1
DFT = 3 -1 =2
SST =
MST = ÷ 2 =

11. MSE = 167.77778 ÷ 24 = 6.9907 HAWKES LEARNING SYSTEMS
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Hypothesis Testing (Two or More Populations)
11.6 ANOVA
Mean Square for Error: =
k = number of treatments = 3
n = number of observations = 27
Degrees of Freedom of Error (DFE) = n - k
DFE = 27-3 =24
SSE =
MSE = ÷ 24 =

12. Reject the Null Hypotheses if p ≤  or
HAWKES LEARNING SYSTEMS
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Hypothesis Testing (Two or More Populations)
11.6 ANOVA =
6.9907
Test Statistic for ANOVA:
0.2066 = =
To determine if the test statistic calculated from the sample is statistically significant we will need to look at the critical value.
The critical values for population variances are found from the chi-square distribution.
Reject the Null Hypotheses if p ≤  or
Reject if F-Statistic (F) ≥ F-Critical (Fdft,dfe)
F-Critical (F2,24) = ( = 0.05)
H0: 1 = 2 = … = n.
Ha: At least one mean differs from the others.

14. HAWKES LEARNING SYSTEMS
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Hypothesis Testing (Two or More Populations)
11.6 ANOVA
ANOVA Table:
Don’t Reject, is NOT ≥ (p-value of is NOT ≤ 0.05)
There is not sufficient evidence at the 0.05 level of significance to support the researcher’s claim that there is a significant difference in the three treatments.

15. SS DF MS F Treatments Error Total n – k
HAWKES LEARNING SYSTEMS
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Hypothesis Testing (Two or More Populations)
11.6 ANOVA
ANOVA Table:
SST = MST x DFT SS DF MS F
Treatments
k – 1
Error
SSE
Total
SST
n – k
SST + SSE
DFT + DFE
DFE = SSE_
MSE

16. SST = MST x (k-1) = x 3 =

17. SSTotal = SST + SSE SSE = SSTotal – SST

18. DFTotal = DFT + DFE
DFE = DFTotal – DFT
DFE = = 11

19. =
293.79
3.86 = =

20. SST

21. SSE

22. DF 1 2 3 4
199.5 19
10.128
9.5521
9.2766
9.1172
7.7086
6.9443
6.5914
6.3882 5
6.6079
5.7861
5.4095
5.1922 6
5.9874
5.1433
4.7571
4.5337 7
5.5914
4.7374
4.3468
4.1203 8
5.3177
4.459
4.0662
3.8379 9
5.1174
4.2565
3.8625
3.6331 10
4.9646
4.1028
3.7083
3.478 11
4.8443
3.9823
3.5874
3.3567 12
4.7472
3.8853
3.4903
3.2592
FIND CRITICAL-F
DFT= 3
DFE= 11
Alpha=
0.05
F-Critical=
=FINV(3,11,0.05)

23. IF 3.8639 > 3.587434, REJECT NULL Ho: All means are equal
HA: At least one mean is different
Reject the Null if F-Statistics (F) > F-Critical
REJECT -- STATISTICALLY SIGNIFICANT
IF > , REJECT NULL

24. Copy Data from Hawkes to Excel
Paste in cell A2

25. MST = SST ÷ DFT
SST = MST x DFT
SST = x 3 =
DFTotal = DFT + DFE
DFE = DFTotal – DFT
DFE = 15 – 3 = 12
F-Statistic = MST ÷ MSE
F-Statistic = ÷
F-Statistic = 0.97
SSTotal = SST + SSE
SSE = SSTotal – SST
SSE = – =